# 3.2.1: How much CO2 is released by burning various fuels

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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)In the Table 3.1. we find how much thermal energy is released by burning a unit weight of various fuels – but an important question also is how much CO_{2} is released in such processes. In fact, it’s easy to find out. The only skill one really needs for doing this is to know how to write the chemical equation of the combustion reaction^{1}.

## How to write the equation of a combustion reaction?

Step One: The left side. On the left side of the equation, you put the chemical formula of a single molecule of the "fuel". For instance: hydrogen occurs in diatomic molecules \(\mathrm{H}_{2}\). Coal, with a not-so-bad approximation, can be though of as pure carbon, C. Natural gas is essentially methane, \(\mathrm{CH}_{4}\). Gasoline is a mixture of a large number of different hydrocarbons, but a "good average" for it may be octane \(\mathrm{C}_{8} \mathrm{H}_{18}\). For methanol and ethanol the respective formulas are \(\mathrm{CH}_{4} \mathrm{O}\) and \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}\). By the way, there are different methods of writing the chemical formulas, some people prefer writing them in a way that is similar to the actual arrangements of atoms in the molecule. Then, methanol is written as \(\mathrm{CH}_{3} \mathrm{OH}\), and ethanol as \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\). But in combustion reaction equations it does not matter which method you prefer: only the total number of atoms of each constituent element must be right. In an ethanol molecule, there are 2 atoms of carbon, 6 atoms of hydrogen, and one atom of oxygen. Note that those numbers are correct, no matter whether you use the \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}\), or the \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) formula.

The other component necessary for combustion is oxygen, usually taken from the air. Oxygen, like hydrogen, also occurs in diatomic molecules \(\mathrm{O}_{2}\).

So, we can now start writing the left side of the combustion equation. Let's take, for example, ethanol. So, we write down the formulas for ethanol and oxygen:

\[ \mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}+x \mathrm{O}_{2} \longrightarrow \]

We don't know yet how many oxygen molecules will be needed, so that we put \(x\).

**Step Two**: The right side. It's even easier, because in the reactions of combustion of all fuels listed in Table 3.1. the only products of combustion may be carbon dioxide \(\mathrm{CO}_{2}\) and water \(\mathrm{H}_{2} \mathrm{O}\). We don't know yet how many molecules of each of the two are created, so we put, as in "ordinary" equations, the symbols normally used for "unknown", \(x, y\) and \(z\) :

\[ \mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}+x \mathrm{O}_{2} \longrightarrow y \mathrm{CO}_{2}+z \mathrm{H}_{2} \mathrm{O} \]

**Step Three**: Determining \(y\) and \(z\). It's really easy. On the right side, we

have a total of 2 carbon atom, so that two molecules of \(\mathrm{CO}_{2}\) can be created and a total of 6 hydrogen atoms, so that three \(\mathrm{H}_{2} \mathrm{O}\) molecules can be created. Therefore:

\[ \mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}+x \mathrm{O}_{2} \longrightarrow 2 \mathrm{CO}_{2}+3 \mathrm{H}_{2} \mathrm{O} \]

**Step Four**: Determining \(x\). We now count the total number of \(O\) atoms on the right side: in two \(\mathrm{CO}_{2}\) molecules there are four of them, and in three \(\mathrm{H}_{2} \mathrm{O}\) molecules there are three of them - seven altogether. And on the left side, one \(\mathrm{O}\) atom already is in \(\mathrm{Hhe} \mathrm{}_{2} \mathrm{H}_{6} \mathrm{O}\) molecule. To get a total of seven on the left side, we need extra six, so that \(x\) must be 3 . So, now we can write the final equation:

\[ \mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}+3 \mathrm{O}_{2} \longrightarrow 2 \mathrm{CO}_{2}+3 \mathrm{H}_{2} \mathrm{O} \]

In a similar fashion, we can write the equation of combustion for methanol:

\[ \mathrm{CH}_{4} \mathrm{O}+1 \frac{1}{2} \mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O} \]

Here the situation is a bit more tricky, because, as you see, on the right side we have four oxygen atoms, and on the left side one is already in the methanol molecule, so we need three extra atoms of oxygen. Some people like to take it this way: three atoms of oxygen are in one-and-one-half oxygen molecule. Other people don't like to use "half-molecules", so they write instead:

\[ 2 \mathrm{CH}_{4} \mathrm{O}+3 \mathrm{O}_{2} \longrightarrow 2 \mathrm{CO}_{2}+4 \mathrm{H}_{2} \mathrm{O} \]

Both ways are correct, and which one you want to use depends only on your personal preferences.

For coal, approximated as carbon, the equation is very simple:

\[ \mathrm{C}+\mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2} \]

For gasoline, the equation is:

\[ \mathrm{C}_{8} \mathrm{H}_{18}+12 \frac{1}{2} \mathrm{O}_{2} \longrightarrow 8 \mathrm{CO}_{2}+9 \mathrm{H}_{2} \mathrm{O} \]

which, if one doesn't like "half-molecules", can also be written as:

\[ 2 \mathrm{C}_{8} \mathrm{H}_{18}+25 \mathrm{O}_{2} \longrightarrow 16 \mathrm{CO}_{2}+18 \mathrm{H}_{2} \mathrm{O} \]

How to calculate how much \(\mathrm{CO}_{2}\) is created in a combustion reaction? To do this, one has to first recall what the atomic mass \(m_{\mathrm{a}}\) is: it's the mass of an atom of a chemical element expressed in atomic mass units. For the hydrogen atom, \(m_{\mathrm{a}}=1\). A carbon atom is 12 times more massive than a hydrogen atom, so that for \(\mathrm{C} m_{\mathrm{a}}=12\), and for an oxygen atom, \(\mathrm{O}\), \(m_{\mathrm{a}}=16\). For simple combustion reactions we consider in this Section, it's all we need.

The molecular mass \(M\) of a molecule is the sum of the atomic masses of all the constituent elements. For instance, for \(\mathrm{CO}_{2}, M=12+2 \times 16=44\). For \(\mathrm{H}_{2} \mathrm{O}\), one gets \(M=2 \times 1+16=18\). For \(\mathrm{O}_{2}, M=2 \times 16=32\). And for ethanol \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}, M=2 \times 12+6 \times 1+16=46\).

In a chemical reaction equation, the sum of all atomic masses on the left side must be equal to the sum on the right side. It's yet another method of checking whether the equation is correct. So, let's look at the combustion equation of ethanol we have derived a moment ago:

\[ \mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}+3 \mathrm{O}_{2} \longrightarrow 2 \mathrm{CO}_{2}+3 \mathrm{H}_{2} \mathrm{O} \]

For the left side, we get: \(\Sigma M_{\text {left }}=46+3 \times 32=142\) ( \(\Sigma\) means "the sum of"). And for the right side, we get: \(\Sigma M_{\text {right }}=2 \times 44+3 \times 18=142\). So that, \(\Sigma M_{\text {left }}=\Sigma M_{\text {right }}\), as expected, showing one more time that our combustion equation is correct.

Now, the calculation of the \(\mathrm{CO}_{2}\) mass becomes very simple. Let's writethe molecular masses instead of the chemical formulas:

\[ \underbrace{46}_{\text {fuel }}+\underbrace{96}_{\text {oxygen }}=\underbrace{88}_{\mathrm{CO}_{2}}+\underbrace{54}_{\mathrm{H}_{2} \mathrm{O}} \]

Now, one can clearly see: the "input" is 46 mass units of ethanol, and the "output" is 88 mass units of \(\mathrm{CO}_{2}\). This gives us the proportion of the output \(\mathrm{CO}_{2}\) mass to the input ethanol mass units (we are not interested in the input oxygen and the output water). There must be the same proportion of masses, if we use kilograms, not atomic mass units. Denote the mass of \(\mathrm{CO}_{2}\) produced by the combustion of \(1 \mathrm{~kg}\) of ethanol as \(X\). We can write:

\[ \frac{\mathrm{CO}_{2} \text { mass (atomic units) }}{\text { Ethanol mass (atomic units) }}=\frac{88}{46}=\frac{X \mathrm{~kg}}{1 \mathrm{~kg}} \]

By solving this proportion, we get:

\[ X=(1 \mathrm{~kg}) \times \frac{88}{46}=1.913 \mathrm{~kg} \]

Ethanol is sometimes used as a fuel - e.g., in tourist stoves - but it's definitely not a fuel used by electric power generating plants. Let's then make similar calculations for two fuels that are used by such plants: coal and natural gas.

Let's take anthracite, the highest-quality mined coal (over \(95 \%\) of pure carbon), and ignore the \(5 \%\) that may not be carbon. The equation of combustion is then pretty simple:

\[ \mathrm{C}+\mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2} \]

The fuel input is 12 atomic mass units, the \(\mathrm{CO}_{2}\) output is 44 atomic mass units. Hence, the combustion of \(1 \mathrm{~kg}\) of anthracite produces:

\[ X=(1 \mathrm{~kg}) \times \frac{44}{12}=3.67 \mathrm{~kg} \text { of } \mathrm{CO}_{2} \]

Natural gas, as mentioned, is almost pure methane \(\mathrm{CH}_{4}\). The equation of combustion is then:

\[ \mathrm{CH}_{4}+2 \mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O} \]

The molecular mass of methane is 16 atomic mass units, so the mass \(X\) of \(\mathrm{CO}_{2}\) created in the reaction is:

\[

X=(1 \mathrm{~kg}) \times \frac{44}{16}=2.75 \mathrm{~kg} \text { of } \mathrm{CO}_{2}

\]

It's only \(3 / 4\) of the \(\mathrm{CO}_{2}\) released by burning the same mass of coal. But one often hears that natural gas is a "much greener" fuel than coal, so perhaps you have expected a more significant difference?

But the difference is significant! The thing is that we should not look at the amount of \(\mathrm{CO}_{2}\) released by burning a mass unit of a given fuel, but at the amount of \(\mathrm{CO}_{2}\) that needs to be released to the atmosphere for obtaining a unit measure - say, \(1 \mathrm{GJ}\) - of thermal energy by burning a givenfuel. So, let's see how such figures look like for coal and natural gas. Let's take the data from Table \(3.1\) and from the calculations we have made, but for 1000 kg (a ton) of fuel, not a single kilogram:

- The burning of \(1000 \mathrm{~kg}\) of coal releases 27.0 GJ of heat, and \(3670 \mathrm{~kg}\) of \(\mathrm{CO}_{2}\). Accordingly, for obtaining \(1 \mathrm{GJ}\) of heat from coal one has to "dump" \(3670 / 27=136 \mathrm{~kg}\) of \(\mathrm{CO}_{2}\) to the atmosphere.
- The burning of \(1000 \mathrm{~kg}\) of natural gas releases \(55 \mathrm{GJ}\) of heat, and 2750 \(\mathrm{kg}\) of \(\mathrm{CO}_{2}\). Therefore, for producing \(1 \mathrm{GW}\) of heat from natural gas one has to send \(2440 / 55=50.0 \mathrm{~kg}\) of \(\mathrm{CO}_{2}\) to the atmosphere.

As follows from the above, natural gas is "nearly three times greener a fuel" than coal! In fact, in practice this ratio may be even higher, because of two reasons. First, we have made our calculations for anthracite, the highest quality coal. It's the so-called smokeless coal \({ }^{2}\). Only about \(1 \%\) of all global coal resources are anthracite, the rest are lesser quality forms with lower heat of combustion than anthracite. A really "bad guy" is lignite, the socalled brown coal. With its heat of combustion as low as \(15 \mathrm{MJ} / \mathrm{kg}\), it may produce as much as \(250 \mathrm{~kg}\) of \(\mathrm{CO}_{2}\) for \(1 \mathrm{GJ}\) of heat, a score about five times worse that natural gas. Nonetheless, "brown coal" is willingly used by some European countries because huge rich deposits of this fuel are placed shallowly underground. Like the coal deposits in Montana – however, in Montana it’s a good quality coal, not the highly polluting lignite.

The other reason that the natural gas is much “greener” than coal is the difference in the efficiency of thermal engines using coal and those using natural gas as fuel. Coal is the standard fuel in power plants using steam turbines – currently, it’s the most popular type of thermal power plants. Here, coal is used to heat steam-generating boilers, from which steam is directed to turbines. The combined efficiency of such a system may be only somewhat higher than 30%. It means that the amount of \(\mathrm{CO}_{2}\) created for producing 1 GW of *electric energy *may be as high as (136 kg)/0.3 ∼ 450 kg of \(\mathrm{CO}_{2}\). On the other hand, the efficiency of *gas turbines*, converting thermal energy from gas burning directly to mechanical energy, may be as high as

45% – and in some newest models of turbines, the record-high efficiency is 62%. Then, the production of 1 GJ of electric power releases only (50.0 kg)/0.62 = 81 kg of \(\mathrm{CO}_{2}\). 81 kg as opposed to 450 kg, natural gas, if such gas turbines are widely used, may be 5.56 times “greener” than coal!

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1 From his past experience in teaching the *Energy Alternatives *course at OSU, the Author knows that not all students taking this course have taken a basic chemistry course, either during their high school education, or at OSU. The term *chemical reaction equation *may sound scary to them – which is a *wrong **reaction*(!), such equations are pretty simple indeed, and we here explain step-by-step how to write them.

2 During the Civil War, the “blockade runners” – i.e., the Confederacy merchant ships trying to sneak through the blockade imposed by the Union – used anthracite as a fuel. Other types of coal were much cheaper, but they produced a lot of smoke, revealing the position of the “blockade runner” to the Union Navy ships enforcing the blockade.